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P(t<X<t+dt)=f(t)dt ?

  1. Sep 2, 2011 #1

    Could anyone show to me that:

    If dt is an infinitely small number, the probability that X is included within the interval (t, t + dt) is equal to f(t) dt ,i.e.
    [itex]Pr(t<X<t+dt) = f(t)dt[/itex]
    where f is the probability density function.

    this sentence is from wikipedia but I could not prove this to myself.

    Thanks if anyone can help.
  2. jcsd
  3. Sep 2, 2011 #2


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    What are the limits of your integral? You can't really make sense of f(t)dt without some kind of constraint on the upper and lower limits.
  4. Sep 2, 2011 #3
    here dt is an infinitely small number. and i don't think the RHS is an integral.

    maybe you can refer to http://en.wikipedia.org/wiki/Probability_density_function#Further_details

    the last line of the Further details
    Last edited: Sep 2, 2011
  5. Sep 2, 2011 #4


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    The probabilty density function, f, is defined by the fact that
    [tex]P(a< x< b)= \int_a^b f(x)dx[/tex]
    or, equivalently,
    [tex]P(a< x< a+h)= \int_a^{a+h} f(x)dx[/tex]

    What Wikipedia gives is a "differential" form of that.
  6. Sep 3, 2011 #5
    Thanks HallsofIvy for your reply.

    I was trying to work out why
    [tex]\int_a^{a+h} f(x)dx= f(a)h[/tex]
    when h infinitely small.

    Could you tell me more about why this is true? or could you explain the differential form? I am still not fully understand yet.
  7. Sep 3, 2011 #6


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    First off, you almost surely do not mean much of what you said literally. You should spend some time thinking about what you actually meant

    Anyways, if by "=" you meant "is approximately" and by "infinitely small" you meant "sufficiently small", it's true because of the mean value theorem and the definition of continuous function.

    (P.S. if f is not assumed to be continuous, then the equation above is very false)
  8. Sep 3, 2011 #7
    It's just a (dodgy) method for turning sums into integrals. In the long run it's worthwhile to learn Stieltjes integration so you can write probabilities and expectations as
    [tex]E[g(X)] = \int_R g(x)dF(x) [/tex]
    which is valid whether or not the distribution F(x) has a density.
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